The total energy loss E in Joules per minute in a magnetic core circuit is partly constant… — JAMB UTME

The total energy loss $E$ in Joules per minute in a magnetic core circuit is partly constant and partly varies directly as the square of the frequency $f$ in Hz of the alternating current. Given that $E = 120\text{ J}$ when $f = 10\text{ Hz}$ , and $E = 360\text{ J}$ when $f = 20\text{ Hz}$ , what is the value of $E$ in Joules when $f = 15\text{ Hz}$ ?

Answer: 220 J

Answer

220

The relation describes a partial variation model

E = c + k f^2

. Substituting the two given conditions (

f=10, E=120

and

f=20, E=360

) yields the system of equations

c + 100k = 120

and

c + 400k = 360

. Solving this system gives

k = 0.8

and

c = 40

. Evaluating

E = 40 + 0.8(15)^2

results in

220\text{ J}

Step-by-Step Solution

Write the general equation for partial variation involving a constant term and a term proportional to

f^2

E = c + k f^2

Partial variation consists of a sum of a constant component and a variable component.

Substitute the known conditions into the variation equation to create a system of linear equations

Equation 1:

c + 100k = 120

; Equation 2:

c + 400k = 360

Two pairs of values are provided to determine the two unknown constants

c

and

k

Solve the system of simultaneous equations for

k

and

c

k = 0.8

and

c = 40

Subtracting Equation 1 from Equation 2 eliminates

c

, allowing direct calculation of

k

, after which

c

is found by substitution.

Calculate the required value of

E

when

f = 15

E = 40 + 0.8(15^2) = 40 + 0.8(225) = 220

Applying the discovered constants to the target frequency yields the final energy loss value.

Key Concept

Partial Variation and Simultaneous Equations

Estimated Time:2m 0s

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